Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:

import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        final Scanner in = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        final int n = in.nextInt();
        final double[] ps = new double[n + 1];
        for (int i = 1; i <= n; i++) {
            ps[i] = in.nextDouble();
        }
        // [win count][current]
        double[][] dp = new double[n + 1][n + 1];
        dp[0][0] = 1;
        for (int i = 1; i <= n; i++) {
            dp[0][i] = dp[0][i - 1] * (1 - ps[i]);
        }

        for (int i = 1; i <= n; i++) {
            for (int j = i; j <= n; j++) {
                double f1 = dp[i - 1][j - 1] * ps[j];
                double f2 = dp[i][j - 1] * (1 - ps[j]);
                dp[i][j] = f1 + f2;
            }
        }
        double result = 0;
        for (int i = n / 2 + 1; i <= n; i++) {
            result += dp[i][n];
        }
        System.out.println(result);
    }
}